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Electrostatics question

2020 · Shift 1 · Q40
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Electrostatics question

2020 · Shift 1 · Q40

JEE AdvancedPhysicsElectrostaticsNumerical+4 / −1
A circular disc of radius R carries surface charge density σ(r)=σ0(1−rR)\sigma \left( r \right) = {\sigma _0}\left( {1 - {r \over R}} \right)σ(r)=σ0​(1−Rr​), where σ\sigmaσ 0 is a constant and r is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is ϕ\phiϕ 0. Electric flux through another spherical surface of radius R4{R \over 4}4R​ and concentric with the disc is ϕ\phiϕ. Then the ratio ϕ0ϕ{{{\phi _0}} \over \phi }ϕϕ0​​ is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 6.4

  1. Use Gauss’s law

For any closed surface,

Φ=Qenclosedε0\Phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0}Φ=ε0​Qenclosed​​

So,

  • ϕ0\phi_0ϕ0​ depends on the total charge on the disc.
  • ϕ\phiϕ depends on the charge enclosed inside the sphere of radius R/4R/4R/4.

Since both spherical surfaces are concentric with the disc, the smaller sphere encloses only that part of the disc lying within radius R/4R/4R/4.

  1. Find total charge on the disc

Given surface charge density:

σ(r)=σ0(1−rR)\sigma(r)=\sigma_0\left(1-\frac{r}{R}\right)σ(r)=σ0​(1−Rr​)

Take a ring element of radius rrr and thickness drdrdr. Its area is

dA=2πr drdA = 2\pi r\,drdA=2πrdr

So charge on the ring is

dq=σ(r) dA=σ0(1−rR)2πr drdq = \sigma(r)\,dA = \sigma_0\left(1-\frac{r}{R}\right)2\pi r\,drdq=σ(r)dA=σ0​(1−Rr​)2πrdr

Hence total charge:

Q0=∫0Rσ0(1−rR)2πr drQ_0 = \int_0^R \sigma_0\left(1-\frac{r}{R}\right)2\pi r\,drQ0​=∫0R​σ0​(1−Rr​)2πrdr Q0=2πσ0∫0R(r−r2R)drQ_0 = 2\pi\sigma_0\int_0^R \left(r-\frac{r^2}{R}\right)drQ0​=2πσ0​∫0R​(r−Rr2​)dr

Now,

∫0Rr dr=R22,∫0Rr2 dr=R33\int_0^R r\,dr = \frac{R^2}{2}, \qquad \int_0^R r^2\,dr = \frac{R^3}{3}∫0R​rdr=2R2​,∫0R​r2dr=3R3​

Therefore,

Q0=2πσ0(R22−1R⋅R33)Q_0 = 2\pi\sigma_0\left(\frac{R^2}{2}-\frac{1}{R}\cdot\frac{R^3}{3}\right)Q0​=2πσ0​(2R2​−R1​⋅3R3​) Q0=2πσ0R2(12−13)Q_0 = 2\pi\sigma_0 R^2\left(\frac12-\frac13\right)Q0​=2πσ0​R2(21​−31​) Q0=2πσ0R2⋅16Q_0 = 2\pi\sigma_0 R^2\cdot\frac16Q0​=2πσ0​R2⋅61​ Q0=πσ0R23Q_0 = \frac{\pi\sigma_0 R^2}{3}Q0​=3πσ0​R2​

Thus,

ϕ0=Q0ε0\phi_0 = \frac{Q_0}{\varepsilon_0}ϕ0​=ε0​Q0​​
  1. Find charge enclosed by sphere of radius R/4R/4R/4

Only the central portion of the disc from r=0r=0r=0 to r=R/4r=R/4r=R/4 lies inside this sphere.

So,

Q=∫0R/4σ0(1−rR)2πr drQ = \int_0^{R/4} \sigma_0\left(1-\frac{r}{R}\right)2\pi r\,drQ=∫0R/4​σ0​(1−Rr​)2πrdr Q=2πσ0∫0R/4(r−r2R)drQ = 2\pi\sigma_0\int_0^{R/4}\left(r-\frac{r^2}{R}\right)drQ=2πσ0​∫0R/4​(r−Rr2​)dr

Now,

∫0R/4r dr=12(R4)2=R232\int_0^{R/4} r\,dr = \frac{1}{2}\left(\frac{R}{4}\right)^2 = \frac{R^2}{32}∫0R/4​rdr=21​(4R​)2=32R2​ ∫0R/4r2 dr=13(R4)3=R3192\int_0^{R/4} r^2\,dr = \frac{1}{3}\left(\frac{R}{4}\right)^3 = \frac{R^3}{192}∫0R/4​r2dr=31​(4R​)3=192R3​

Hence,

Q=2πσ0(R232−1R⋅R3192)Q = 2\pi\sigma_0\left(\frac{R^2}{32}-\frac{1}{R}\cdot\frac{R^3}{192}\right)Q=2πσ0​(32R2​−R1​⋅192R3​) Q=2πσ0R2(132−1192)Q = 2\pi\sigma_0 R^2\left(\frac{1}{32}-\frac{1}{192}\right)Q=2πσ0​R2(321​−1921​) Q=2πσ0R2(6−1192)Q = 2\pi\sigma_0 R^2\left(\frac{6-1}{192}\right)Q=2πσ0​R2(1926−1​) Q=2πσ0R2⋅5192Q = 2\pi\sigma_0 R^2\cdot\frac{5}{192}Q=2πσ0​R2⋅1925​ Q=5πσ0R296Q = \frac{5\pi\sigma_0 R^2}{96}Q=965πσ0​R2​

Thus,

ϕ=Qε0\phi = \frac{Q}{\varepsilon_0}ϕ=ε0​Q​
  1. Find the ratio
ϕ0ϕ=Q0Q\frac{\phi_0}{\phi} = \frac{Q_0}{Q}ϕϕ0​​=QQ0​​ ϕ0ϕ=πσ0R235πσ0R296\frac{\phi_0}{\phi} = \frac{\frac{\pi\sigma_0 R^2}{3}}{\frac{5\pi\sigma_0 R^2}{96}}ϕϕ0​​=965πσ0​R2​3πσ0​R2​​

Cancel common terms:

ϕ0ϕ=13⋅965\frac{\phi_0}{\phi} = \frac{1}{3}\cdot\frac{96}{5}ϕϕ0​​=31​⋅596​ ϕ0ϕ=325=6.4\frac{\phi_0}{\phi} = \frac{32}{5} = 6.4ϕϕ0​​=532​=6.4
  1. Final answer
ϕ0ϕ=325=6.4\boxed{\frac{\phi_0}{\phi}=\frac{32}{5}=6.4}ϕϕ0​​=532​=6.4​

The derived answer matches the stored correct answer.

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